Question

Find the value of

$f ( t ) = \sum_{k = 1}^{n} \begin{vmatrix} \sin ( 2 K t ) & \cos ( ( n + 1 ) t ) & \cos ( n t ) \\ \cos ( ( n + 1 ) t ) & \cos ( n t ) & \cos t \\ \sin ( ( n + 1 ) t ) & \sin ( n t ) & \sin t \end{vmatrix}$

Answer 1

$-\frac{\sin((n-1)t)\sin(nt)\sin((n+1)t)}{\sin(t)} + n(\cos((n+1)t)\sin(nt) - \cos(nt)\sin(t))$

Answer 2

To find the value of $f(t)$, we first need to evaluate the determinant: $D_k = \begin{vmatrix} \sin(2Kt) & \cos((n+1)t) & \cos(nt) \\ \cos((n+1)t) & \cos(nt) & \cos(t) \\ \sin((n+1)t) & \sin(nt) & \sin(t) \end{vmatrix}$

Let $X = (n+1)t$, $Y = nt$, $Z = t$. The determinant can be written as: $D_k = \begin{vmatrix} \sin(2Kt) & \cos X & \cos Y \\ \cos X & \cos Y & \cos Z \\ \sin X & \sin Y & \sin Z \end{vmatrix}$

Expand the determinant along the first row: $D_k = \sin(2Kt) (\cos Y \sin Z - \cos Z \sin Y) - \cos X (\cos X \sin Z - \cos Z \sin X) + \cos Y (\cos X \sin Y - \cos Y \sin X)$

Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $\cos Y \sin Z - \cos Z \sin Y = -(\sin Y \cos Z - \cos Y \sin Z) = -\sin(Y-Z)$ $\cos X \sin Z - \cos Z \sin X = -(\sin X \cos Z - \cos X \sin Z) = -\sin(X-Z)$ $\cos X \sin Y - \cos Y \sin X = -(\sin X \cos Y - \cos X \sin Y) = -\sin(X-Y)$

Substitute back the values of $X, Y, Z$: $Y-Z = nt - t = (n-1)t$ $X-Z = (n+1)t - t = nt$ $X-Y = (n+1)t - nt = t$

So, the determinant becomes: $D_k = \sin(2Kt) (-\sin((n-1)t)) - \cos((n+1)t) (-\sin(nt)) + \cos(nt) (-\sin(t))$ $D_k = -\sin(2Kt)\sin((n-1)t) + \cos((n+1)t)\sin(nt) - \cos(nt)\sin(t)$

Now, we need to find $f(t) = \sum_{k=1}^{n} D_k$. $f(t) = \sum_{k=1}^{n} [-\sin(2Kt)\sin((n-1)t) + \cos((n+1)t)\sin(nt) - \cos(nt)\sin(t)]$

The terms $\cos((n+1)t)\sin(nt) - \cos(nt)\sin(t)$ are independent of $k$.

So, $f(t) = -\sin((n-1)t) \sum_{k=1}^{n} \sin(2Kt) + n[\cos((n+1)t)\sin(nt) - \cos(nt)\sin(t)]$

The sum of sines in arithmetic progression is given by $\sum_{k=1}^{n} \sin(kx) = \frac{\sin(nx/2)\sin((n+1)x/2)}{\sin(x/2)}$. Here, $x = 2t$. $\sum_{k=1}^{n} \sin(2Kt) = \frac{\sin(n(2t)/2)\sin((n+1)(2t)/2)}{\sin(2t/2)} = \frac{\sin(nt)\sin((n+1)t)}{\sin(t)}$

Substitute this back into the expression for $f(t)$: $f(t) = -\sin((n-1)t) \frac{\sin(nt)\sin((n+1)t)}{\sin(t)} + n[\cos((n+1)t)\sin(nt) - \cos(nt)\sin(t)]$